# Question #c7855

Feb 9, 2014

To find which side of a reversible reaction has more moles, you add up the coefficients on each side of the equation and compare the numbers.

The number of moles on each side of a reversible reaction is important when you are trying to determine the effect of pressure on a position of equilibrium. A change in pressure will change the volume, but only for gases. Thus, you consider only the moles of gases on each side of the equation. You ignore any solids and liquids, because their volumes do not change.

Examples

A(g) + 2B(g) ⇌ C(g) + D(g); 3 mol gas on left; 2 mol gas on right;
∴ more moles on left.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g); 4 mol gas on left; 2 mol gas on right;
∴ more moles on left.

2NOCl(g) ⇌ 2NO(g) + Cl₂(g); 2 mol gas on left; 3 mol gas on right; ∴ more moles on right.

C(s) + 2H₂(g) ⇌ CH₄(g); 2 mol gas on left; 1 mol gas on right;
∴ more moles on left.

MgCO₃(s) ⇌ MgO(s) + CO₂(g); 0 mol gas on left; 1 mol gas on right; ∴ more moles on right.

Remember: For the effect of pressure on a reversible reaction, count only the moles of gases.